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SCORPION-xisa [38]
3 years ago
11

ਨਿਰਭਰ ਕਰਦੀਇਬਾਣੀ ਸਰੀਰਜ ਸਮੱਰਥਾ ਜਿਹੜਾ ਕਿੰ- ਉਨ ਬਾਦ ਤੇ​

Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

I can't understand

Explanation:

But if you have a english translator I can help

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675 km equals how much cm ?
irga5000 [103]
KHDMDCM.
Now go from Kilometer to Centimeter: 5.
Move the decimal 5 places to the right: 67,500,000 centimeters.
Hope this helps :)
7 0
3 years ago
Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no
marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s. 

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

8 0
3 years ago
Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric fiel
ad-work [718]

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

5 0
3 years ago
Fast and safe heart rate for workouts is called muscular strength? True or false
Vsevolod [243]

Answer:

False

Explanation:

8 0
2 years ago
Read 2 more answers
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080
deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}

now we have

M = 0.120 kg

m_1 = 0.02 kg

m_3 = 0.08 kg

now we have

I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}

so we have

I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162

I = 0.02835

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}

\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)

\omega = 4.32 rad/s

Now speed of 0.08 kg mass when it reaches to bottom point is given as

v = \omega \frac{L}{2}

v = 4.32 (0.45)

v = 1.94 m/s

3 0
3 years ago
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